NCERT Exemplar - MCQs

Chapter 6 Class 12 Application of Derivatives
Serial order wise

## (C) (2, 0)         (D) (0, 2)

### Transcript

Question 12 The tangent to the curve γπ¦=πγ^2π₯ at the point (0, 1) meets x-axis at: (0, 1) (B) β 1/2, 0 (C) (2, 0) (D) (0, 2) Given γπ¦=πγ^2π₯ Finding slope of tangent at (0, 1) γπ¦=πγ^2π₯ Differentiating w.r.t. π₯ ππ/ππ = 2π^ππ At point (0, 1) Putting π₯=0, π¦=0 in ππ¦/ππ₯ ππ¦/ππ₯ = γ2πγ^0 ππ¦/ππ₯ = 2 Γ 1 ππ/ππ=π Finding equation of tangent Equation of line at (π₯1 , π¦1) & having Slope m is π¦βπ¦1=π(π₯βπ₯1) β΄ Equation of tangent at point (0, 1) & having slope 2 is (πβπ) = 2 (πβπ) (π¦β1) = 2π₯ π=ππ+π Now, we need to find the point where tangent meets the x-axis Since point is on the x-axis, itβs y-coordinate will be 0 β΄ Point = (π,π) Putting π¦ = 0 in (1) 0 = 2π + 1 β1 = 2π₯ (β1)/2=π₯ π = (βπ)/π Thus, required point is ((βπ)/π, π) So, the Correct answer is (B)